WAEC SSCE - Further Mathematics - 2021

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Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

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To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

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The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

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To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

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To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

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Detalles de la respuesta

To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

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To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

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Detalles de la respuesta

To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

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A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

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Detalles de la respuesta

To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

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Detalles de la respuesta

The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

Detalles de la respuesta

(a) To find the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x² - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.

Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x² - x + 1)(x - 2) with respect to x, giving:

dy/dx = 3x² - 8x + 1

Substituting x = 2, we get:

dy/dx = 3(2)² - 8(2) + 1 = -7

Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.

Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.

Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:

y - 0 = (1/7)(x - 2)

Simplifying this equation gives:

y = (1/7)x - 2/7

So the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.

(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).

Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:

m = (y₂ - y₁)/(x₂ - x₁)

Substituting the coordinates of Q and the midpoint of PR gives:

m = (1 - (-1))/(5 - 1) = 1/2

Detalles de la respuesta

(a) Draw a Histogram

Given the data on monthly income distribution, we can create a histogram. Here's a step-by-step process for drawing the histogram:

1. Class Intervals: The monthly income classes are given as ranges (in thousands of naira).
2. Frequency: The number of workers in each class interval represents the frequency.

Let's list the class intervals and their corresponding frequencies:

• 135-139: 20 workers
• 140-149: 42 workers
• 150-154: 28 workers
• 155-164: 38 workers
• 165-169: 22 workers

We'll plot these intervals on the x-axis and the number of workers on the y-axis.

(b) Estimate the Mode

The mode of a distribution is the value that appears most frequently. In a histogram, the mode corresponds to the class interval with the highest frequency.

1. Identify the class interval with the highest frequency.

2. Use the following formula for estimating the mode in a grouped frequency distribution:

   $\text{Mode}=L+\left(\frac{(f_m?f_1)}{(2f_m?f_1?f_2)}\right)\times w\backslash text\{Mode\}\; =\; L\; +\; \backslash left(\; \backslash frac\{(f\_m\; -\; f\_\{1\})\}\{(2f\_m\; -\; f\_\{1\}\; -\; f\_\{2\})\}\; \backslash right)\; \backslash times\; w$

   where:

   • $LL$ is the lower boundary of the modal class,
   • $f_{m}f\_m$ is the frequency of the modal class,
   • $f_{1}f\_\{1\}$ is the frequency of the class preceding the modal class,
   • $f_{2}f\_\{2\}$ is the frequency of the class succeeding the modal class,
   • $ww$ is the width of the class intervals.

Let's plot the histogram and estimate the mode.

Drawing the Histogram and Estimating the Mode

(a) Histogram

The histogram above displays the distribution of monthly income (in thousands of naira) among workers in a factory. The x-axis represents the monthly income ranges, while the y-axis shows the number of workers in each range.

(b) Estimation of the Mode

Using the histogram and the data provided, the mode can be estimated with the following steps:

1. Identify the Modal Class:

   • The class interval with the highest frequency is $140?149140-149$ with 42 workers.

2. Apply the Mode Formula:

   • $L=140L\; =\; 140$ (lower boundary of the modal class)
   • $f_m=42f\_m\; =\; 42$ (frequency of the modal class)
   • $f_1=20f\_1\; =\; 20$ (frequency of the class preceding the modal class, which is $135?139135-139$)
   • $f_2=28f\_2\; =\; 28$ (frequency of the class succeeding the modal class, which is $150?154150-154$)
   • $w=10w\; =\; 10$ (width of the class intervals)

   Using the formula for the mode:

   $\text{Mode}=140+\left(\frac{(42?20)}{(2\times 42?20?28)}\right)\times 10\backslash text\{Mode\}\; =\; 140\; +\; \backslash left(\; \backslash frac\{(42\; -\; 20)\}\{(2\; \backslash times\; 42\; -\; 20\; -\; 28)\}\; \backslash right)\; \backslash times\; 10$

   Simplifying this:

   $\text{Mode}=140+\left(\frac{22}{84?48}\right)\times 10\backslash text\{Mode\}\; =\; 140\; +\; \backslash left(\; \backslash frac\{22\}\{84\; -\; 48\}\; \backslash right)\; \backslash times\; 10$ $\text{Mode}=140+\left(\frac{22}{36}\right)\times 10\backslash text\{Mode\}\; =\; 140\; +\; \backslash left(\; \backslash frac\{22\}\{36\}\; \backslash right)\; \backslash times\; 10$ $\text{Mode}?140+6.111\backslash text\{Mode\}\; \backslash approx\; 140\; +\; 6.111$ $\text{Mode}?146.11\backslash text\{Mode\}\; \backslash approx\; 146.11$

Therefore, the estimated mode of the distribution is approximately 146.11 thousand naira.

Detalles de la respuesta

(a) (i) To construct a cumulative frequency table, we need to add up the frequencies for each height interval and all the intervals before it. The cumulative frequency for each interval is shown in the table below:

(a) (ii) To draw a cumulative frequency curve, we plot the cumulative frequency for each interval against the upper limit of the interval. We then join the points with straight lines to form the curve, as shown below:

(b) (i) To estimate the median height, we locate the value on the cumulative frequency curve that corresponds to the 50th percentile. This is the point where the curve intersects the line at y = 0.5. We draw a line down to the x-axis to find the corresponding height. From the graph, we can see that the median height is around 124-125cm.

(b) (ii) To estimate the interquartile range (IQR), we need to find the heights corresponding to the 25th and 75th percentiles on the cumulative frequency curve. We draw lines down from these points to the x-axis to find the corresponding heights. The IQR is then the difference between these heights. From the graph, we can estimate the 25th percentile at around 116-117cm and the 75th percentile at around 134-135cm. Therefore, the IQR is around 18-19cm.

(b) (iii) To estimate the percentage of students whose heights are most 130cm, we locate the point on the cumulative frequency curve that corresponds to 130cm. We then draw a line across to the y-axis to find the corresponding percentile. From the graph, we can see that the percentage of students whose heights are most 130cm is around 60-65%.

Detalles de la respuesta

9∫1 ${}^{}{<br>}_{}$ x(2x−3)√x $\frac{<br>}{}$ dx = 2x2−3x)x1/2 $\frac{{\mathrm{<br>}}^{}<br>}{{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"><br></span>}}^{}}$

= x −1/2 ${}^{ \; \; \; }$ (2x2 ${}^2$ - 3x)

= 2x3/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$ - 3x1/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$

= 2x3/23/2+1 $\frac{{\mathrm{<br>}}^{\mathrm{<br>}}}{}$ - 3x1/21/2+1 $\frac{{\mathrm{<br>}}^{}}{\mathrm{<br>}}$

= 2x3/25/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 3x1/23/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$

= 9∫1 ${}^{}{<br>}_{}$ [4x3/25 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 6x1/23 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$] 

[ 4(9)3/25 $\frac{<br>}{}$- 6(9)1/23 $\frac{{<br>}^{}}{}$] - [ 4(9)1/25 $\frac{<br>}{}$ - 6(1)1/23 $\frac{<br>}{}$]

= [ 4(27)5 $\frac{<br>}{}$- 6(3)3 $\frac{<br>}{}$- [ 45 - 2 ]

=  [ 1085- 2 ] -  [ 65 ]

= 785 + 65= 845 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br><br></span>}}{}$

164/5

Detalles de la respuesta

Total bulbs:

n(Red)=5, n(B) = 7, n(G) = 4

(5+7+4) = 16

p(R)= 5/16, p(B) = 7/16, p(G) = 4/16

(a) p(All same colour of bulbs) 

= p(RR) Or p(BB) or p(GG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 716 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 615 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 416 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 315

= 112 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 740 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

=  10+21+6120 $\frac{<br>\mathrm{<br>}}{}$ =  37120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

 (b) All different colors = 1-p (AIl the same colour) =

1 - 37120 $\frac{37}{120}$ = 83120 $\frac{83}{120}$

(c) p(At least one red)  = p(RR) + p(RB) + p(RG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 715 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

= 112 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 748 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  112 $\frac{\mathrm{<br>}}{}$

8+748 $\frac{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}{\mathrm{<br>}}$ = 1548 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

= 516

Detalles de la respuesta

(a)

Time taken to reach the ground

Using S = ut + 1/2gt2 ${}^2$

 (u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t2 ${}^2$

5t2 ${}^2$ = 50;

t2 ${}^2$ =50/5 = 10

 t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b)

Using Lami's rule,

20sin34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = T1sin90 $\frac{{\mathrm{<br>\; <br>}}_{}}{\mathrm{<br>}}$ = T2sin56 $\frac{{\mathrm{<br>\; <br>}}_{}}{}$

= T1 ${}_1$ 20.sin90sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ = T2 ${}_2$ 20.sin56sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ 

T1 ${}_1$ = 35.79N and T2 ${}_2$ = 29.65N